Archive for the ‘Trigonography’ Category

Cotangent Identity

$$\begin{array}{cc} & |\overline{AC}|\,|\overline{FD}| = |AEDF| = |\overline{AB}|\,|\overline{ED}| \\[6pt] \to \qquad & \large{b^2\,\left(\,\cot A + \cot B\,\right) = c^2\,\left(\,\cot A + \cot C\,\right)} \\ \end{array}$$

Posted 23 September, 2015 by Blue in Proof Without Words, Trigonography

Half-Angles in a Triangle

My answer to this question at $$\Large{c\;\sin\frac{A-B}{2} \;\equiv\; (a-b)\;\cos\frac{C}{2}}$$

Posted 12 August, 2015 by Blue in Proof Without Words, Trigonography

Identity from

Taken from my answer to “Proving a weird trig identity” at . $$\Large{\frac{\cos\theta}{1-\sin\theta} = \frac{\sec\theta + \tan\theta}{1}}$$ In the diagram, \(\overline{AB}\) is tangent to the unit circle at P. The “trig lengths” (except for \(|\overline{AQ}|\)) should be clear. We note that \(\angle BPR \cong \angle RPP^\prime\), since these inscribed angles subtend congruent arcs \(\stackrel{\frown}{PR}\) […]

Posted 14 May, 2015 by Blue in Classroom, Trigonography

Product Identities

$$\begin{align} 2 \cos A \cos B &= \cos(A-B) \;+\; \cos(A+B) \\ 2 \sin A \;\sin B &= \cos(A-B) \;-\; \cos(A+B) \end{align}$$

Posted 25 February, 2015 by Blue in Classroom, Proof Without Words, Trigonography, Uncategorized

Proof Without Words: Nine-Point Circle Property

The “Nine-Point Circle” of a triangle \(\triangle ABC\) contains the vertices of the medial triangle (\(\triangle RST \)) and orthic triangle (\(\triangle UVW \)), as well as points \(X\), \(Y\), \(Z\) that bisect segments from \(\triangle ABC\)’s orthocenter (\(H\)) to its vertices. A question on Math.StackExchange.comĀ asked for proof that the center of the nine-point circle […]